Problem: A goblet contains $5$ red balls, $6$ green balls, and $3$ blue balls. If we choose a ball, then another ball without putting the first one back in the goblet, what is the probability that the first ball will be blue and the second will be green? Write your answer as a simplified fraction.
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a blue ball and leaving it out. Event B is picking a green ball. Let's take the events one at at time. What is the probability that the first ball chosen will be blue? There are $3$ blue balls, and $14$ total, so the probability we will pick a blue ball is $\dfrac{3} {14}$ After we take out the first ball, we don't put it back in, so there are only $13$ balls left. Since the first ball was blue, there are still $6$ green balls left. So, the probability of picking a green ball after taking out a blue ball is $\dfrac{6} {13}$ Therefore, the probability of picking a blue ball, then a green ball is $\dfrac{3}{14} \cdot \dfrac{6}{13} = \dfrac{9}{91}$